JEE Previous Year Questions on Definite Integrals - Download free pdf

JEE Previous year questions on the topic definite integrals are available here. Before we start solving problems, let us have a brief look on, what is definite integrals? “A definite integral is an integral with lower and upper limits”. This article covers definite integral questions from the past year of JEE Main along with the detailed solution for each question. These questions include all the important topics and formulae. Students can also increase their problem-solving efficiency by referring to the solved examples and practising them. About 2-4 questions are asked from this topic in JEE Examination every year. Aspirant can download free pdf and practice all the questions and get ready for the exam.

JEE Main Definite Integration Past Year Questions With Solutions

Question 1: If n is any integer, then find the value of

\begin{array}{l} \int_{0}^{π} e^{\cos^{2} x} \cos^{3} (2 n + 1) x d x \end{array}

Solution:

Since

\begin{array}{l} \cos (2 n + 1) (π - x) = \cos [(2 n + 1) π - (2 n + 1) x] \\ = - \cos (2 n + 1) x and \cos^{2} (π - x) \end{array}

\begin{array}{l} = \cos^{2} x \end{array}

So that f (2a − x) = −f (x), and hence by the property of definite integral,

\begin{array}{l} \int_{0}^{π} e^{\cos^{2} x} \cos^{3} (2 n + 1) x d x = 0 . \end{array}

Question 2: Let f be a positive function. Let

\begin{array}{l} I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x \end{array}

\begin{array}{l} I_{2} = \int_{1 - k}^{k} f {x (1 - x)} d x \end{array}

when 2k − 1 > 0. Then find I₁/I₂.

Solution:

\begin{array}{l} I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x \\ = \int_{1 - k}^{k} (1 - k + k - x) f [(1 - k + k - x) {1 - (1 - k + k - x)}] d x \end{array}

\begin{array}{l} (∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x) \\ = \int_{1 - k}^{k} (1 - x) f {x (1 - x)} d x \\ = \int_{1 - k}^{k} f {x (1 - x)} d x - \int_{1 - k}^{k} x f {x (1 - x)} d x = I_{2} - I_{1} \\ ∴ 2 I_{1} = I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{2} \end{array}

Question 3: Let a, b, c be non-zero real numbers such that

\begin{array}{l} \int_{0}^{1} (1 + \cos^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + \cos^{8} x) (a x^{2} + b x + c) d x \end{array}

Then the quadratic equation ax² + bx + c = 0 has

(a) no root in (0, 2)

(b) at least one root in (1, 2)

(c) a double root in (0, 2)

(d) two imaginary roots

Solution:

f(x) = ∫₀ ^x (1 + cos⁸x)(ax²+ bx + c) dx

f(1) = f(2)

By Rolle’s theorem, there exist at least one point k ∈ (1, 2) such that f'(k) = 0

f’(x) = (1 + cos⁸x)(ax² + bx + c)

f’(k) = 0

(1 + cos⁸k)(ak² + bk + c) = 0

ak² + bk + c = 0 (as 1 + cos⁸k ≠ 0)

So x = k is a root of ak² + bk + c = 0

k ∈ (1, 2)

So atleast one root in (1, 2).

Hence option b is the answer.

Question 4: The value of

\begin{array}{l} \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x, a > 0, \end{array}

is

Solution:

\begin{array}{l} I = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{x}} d x = \int_{π}^{- π} \frac{\cos^{2} x}{1 + a^{- x}} (- d x) \\ = \int_{- π}^{π} \frac{\cos^{2} x}{1 + a^{- x}} d x \\ \Rightarrow I + I = \int_{- π}^{π} \cos^{2} x (\frac{1}{1 + a^{x}} + \frac{1}{1 + a^{- x}}) d x \\ = \int_{- π}^{π} \cos^{2} x d x \\ 2 I = 2 \int_{0}^{π} \cos^{2} x . d x = \int_{0}^{π} (1 + \cos 2 x) d x \\ 2 I = [x]_{0}^{π} + {[\frac{\sin 2 x}{2}]}_{0}^{π} \\ \Rightarrow 2 I = π \Rightarrow I = \frac{π}{2} . \end{array}

Question 5: If

\begin{array}{l} l (m, n) = \int_{0}^{1} t^{m} {(1 + t)}^{n} d t, \end{array}

then the expression for l (m, n) in terms of l (m + 1, n − 1) is

Solution:

\begin{array}{l} l (m, n) = \int_{0}^{1} t^{m} (1 + t)^{n} d t \\ {[{(1 + t)}^{n} \frac{t^{m + 1}}{m + 1}]}_{0}^{1} - \int_{0}^{1} n {(1 + t)}^{n - 1} \frac{t^{m + 1}}{m + 1} d t \\ = \frac{2^{n}}{m + 1} - \frac{n}{m + 1} l (m + 1, n - 1) . \end{array}

Question 6: The area bounded by the curves y = |x| − 1 and y = −|x| + 1 is

Solution:

The lines are y = x − 1 , x ≥ 0

y = −x − 1, x < 0, y = −x + 1, x ≥ 0, y = x + 1, x < 0

Area

\begin{array}{l} = 4 \times (\frac{1}{2} \times 1 \times 1) = 2 \end{array}

Question 7: If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral

\begin{array}{l} \int_{π / 2}^{3 π / 2} [2 \sin x] d x \end{array}

is

Solution:

We know

\begin{array}{l} - 1 \leq \sin x \leq 1 \Rightarrow - 2 \leq 2 \sin x \leq 2 \\ I = \int_{\frac{π}{2}}^{\frac{3 π}{2}} [2 \sin x] d x \\ = \int_{\frac{π}{2}}^{\frac{5 π}{6}} [2 \sin x] d x + \int_{\frac{5 π}{6}}^{π} [2 \sin x] d x + \int_{π}^{\frac{7 π}{6}} [2 \sin x] d x + \int_{\frac{7 π}{6}}^{\frac{3 π}{2}} [2 \sin x] d x \\ = \int_{\frac{π}{2}}^{\frac{5 π}{6}} (1) d x + \int_{\frac{5 π}{6}}^{π} (0) d x + \int_{π}^{\frac{7 π}{6}} (- 1) d x + \int_{\frac{7 π}{6}}^{\frac{3 π}{2}} (- 2) d x \\ = (\frac{5 π}{6} - \frac{π}{2}) + 0 - (\frac{7 π}{6} - π) - 2 (\frac{3 π}{2} - \frac{7 π}{6}) \\ = \frac{2 π}{6} - \frac{π}{6} - \frac{4 π}{6} = - \frac{π}{2} . \end{array}

Question 8: If

\begin{array}{l} f (x) = A \sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2} and \int_{0}^{1} f (x) d x = \frac{2 A}{π}, \end{array}

then find the constants A and B respectively.

Solution:

\begin{array}{l} f (x) = A \sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2} \\ \int_{0}^{1} f (x) d x = \frac{2 A}{π}, \\ \int_{0}^{1} {A \sin (\frac{π x}{2}) + B} d x = \frac{2 A}{π} \\ {| - \frac{2 A}{π} \cos \frac{π x}{2} + B x |}_{0}^{1} = \frac{2 A}{π} \\ B - (\frac{- 2 A}{π}) = \frac{2 A}{π} \Rightarrow B = 0 \\ ∴ f (x) = A \sin \frac{π x}{2} \Rightarrow f^{'} (x) = \frac{π A}{2} \cos \frac{π x}{2} \\ and f^{'} (\frac{1}{2}) = \frac{π A}{2} (\frac{1}{\sqrt{2}}) = \sqrt{2} \Rightarrow π A = 4 \Rightarrow A = \frac{4}{π} \\ A = \frac{4}{π}, B = 0. \end{array}

Question 9: Evaluate

\begin{array}{l} \int_{0}^{π} \frac{\sin (n + \frac{1}{2}) x}{\sin (x / 2)} d x, (n \in N) \end{array}

Solution:

\begin{array}{l} 2 \sin \frac{x}{2} . (\frac{1}{2} + \cos x + \cos 2 x + \dots . . + \cos n x) \\ = \sin \frac{x}{2} + 2 \sin \frac{x}{2} \cos x + 2 \sin \frac{x}{2} \cos 2 x + \dots . + 2 \sin \frac{x}{2} \cos n x \\ = \sin \frac{x}{2} + \sin \frac{3 x}{2} - \sin \frac{x}{2} + \sin \frac{5 x}{2} - \sin \frac{3 x}{2} + \dots . . + \sin (n + \frac{1}{2}) x - \sin (n - \frac{1}{2}) x \\ = \sin (n + \frac{1}{2}) x \\ \frac{1}{2} + \cos x + \cos 2 x + \dots . . + \cos n x = \frac{\sin (n + \frac{1}{2}) x}{2 \sin (\frac{x}{2})} \\ \int_{0}^{π} \frac{\sin (n + \frac{1}{2}) x}{\sin (\frac{x}{2})} d x = 2 (\int_{0}^{π} \frac{1}{2} d x + \int_{0}^{π} \cos x d x + \dots . . + \int_{0}^{π} \cos n x d x) \\ = 2 {(\frac{π}{2} + \sin x + \dots . . + \frac{\sin n x}{n})}_{0}^{π} = π . \end{array}

Question 10: The sine and cosine curves intersect infinitely many times giving bounded regions of equal areas. Then find the area of one such region.

Solution:

Point of intersection of y = sin x and y = cos x is π/4, 3π/4, 5π/4.

Since, sin x ≥ cos x on the interval [π/4, 3π/4].

Therefore, area of one such region

\begin{array}{l} = \int_{π / 4}^{3 π / 4} (\sin x - \cos x) d x \end{array}

= 2/√2

= √2

Question 11: The integral

\begin{array}{l} \int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 c o s 2 x}{(t a n x + c o t x)^{3}} d x \end{array}

equals

(1) 15/18 (2) 13/32 (3) 13/256 (4) 15/64

Answer: (1)

Solution:

Let

\begin{array}{l} I = \int_{\frac{π}{12}}^{\frac{π}{4}} \frac{8 c o s 2 x}{(t a n x + c o t x)^{3}} d x \end{array}

JEE Past Year Questions on Definite Integration

\begin{array}{l} I = \int_{\frac{π}{12}}^{\frac{π}{4}} 8 c o s 2 x \times \frac{s i n^{3} (2 x)}{8} d x \end{array}

\begin{array}{l} I = \int_{\frac{π}{12}}^{\frac{π}{4}} (s i n 2 x c o s 2 x) \times s i n^{2} (2 x) d x \end{array}

\begin{array}{l} I = \frac{1}{2} \int_{\frac{π}{12}}^{\frac{π}{4}} s i n 4 x (\frac{1 - c o s 4 x}{2}) d x \end{array}

\begin{array}{l} I = \frac{1}{4} \int_{\frac{π}{12}}^{\frac{π}{4}} s i n 4 x d x - \frac{1}{8} \int_{\frac{π}{12}}^{\frac{π}{4}} s i n 8 x d x \end{array}

On solving the above integral and putting lower and upper limits, we have

I = 15/128 (Answer!)

Question 12: If f(a + b + 1 – x) = f(x) ∀ x, where a and b are fixed positive real numbers, then the below expression is equal to

JEE Past Year Questions on Definite Integrals

Solution:

f(a + b + 1 – x) = f(x) …(1)

At x -> x + 1

f(a + b – x) = f(x+1) …(2)

From (1) and (2),

\begin{array}{l} I = \frac{1}{a + b} \int_{a}^{b} (a + b - x) [f (x + 1) + f (x)] d x \dots . (4) \end{array}

Adding equation (4) to given, we have

Definite Integrals Past Year Solved Questions

Question 13: Find the value of

\begin{array}{l} \int_{0}^{1} \frac{8 l o g (1 + x)}{1 + x^{2}} d x \end{array}

Solution:

\begin{array}{l} \int_{0}^{1} \frac{8 l o g (1 + x)}{1 + x^{2}} d x \end{array}

Put x = tanθ, So θ = tan^-1 x

dθ = 1/(1+x²) dx

Now, let

\begin{array}{l} I = \int_{0}^{\frac{π}{4}} 8 l o g (1 + t a n θ) d θ \dots . (1) \end{array}

Lets say, θ = π/4 – θ because x + y = π/4

⇒ (1 + tan x)(1 + tan y) = 2

(1) ⇒ I = π log 2

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